In my last two posts, I introduced you to some of the basics of combinatorics: permutations and combinations. Now, I’m going to tie this concept into something that Jeopardy! tournament players might find useful:

How do I calculate the probabilities for wild card contenders to advance?

With the help of several sources – J! Archive, JBoard.tv, Kelly the Wagering Strategist’s notes – I’ve compiled most of the past non-winning scores from quarterfinals. (You can find a table with all historical ToC quarterfinal scores at the bottom of this post.)

For my model, I take those scores as roughly representative of what might happen in a given tournament. Using those numbers, it’s just a matter of calculating the probability that a given score gets bumped off the leaderboard.

In this post, we’ll look at the situation after the second quarterfinal of the most recent Tournament of Champions, won by Alex Jacob. For those who have already played, I’m including the percentage of past non-winning ToC scores that finished below their totals.

Note that I do not include the winners of future quarterfinals. In any given game, at most two players will have a shot at a wild card; I’m using the faces you see here to make the examples easier to follow.

The Bubble Player

The easiest odds to calculate are those of the player sitting “on the bubble”– in the fourth wild-card slot. If just one non-winner finishes with a higher total, the Bubble Player’s tournament is over.*

After two games, Greg Seroka sat in the fourth position. His total of 11,500 was better than 71% of all past non-winning scores. As I noted at the time, 11,500 is unusually high for that spot with so many games yet to be played; the cut-off in the 2014 ToC, after all, was 9,100.

Using this number as a probability, what are the odds that none of the four remaining non-winners beat Greg? (Remember, when we want to determine the probability that ALL independent events happen, we multiply their probabilities together.)

*Yes, a fifth wild-card slot will open up if a QF ends with no winner. That possibility has almost no effect on these calculations.

The #3 spot: slightly better

The higher positions are harder to calculate, because there are more ways that can work out in those players’ favor. Let’s consider the third-place position, held by Kristin Sausville. She had 12,800, better than 78% of past non-winning scores.

What is the probability none of the remaining players beats her?

78% x 78% x 78% x 78% x 78% x 78% = 0.786 = 22.5%

Not so great, either. But Kristin has a bit more wiggle room than Greg: it’s fine if one non-winner beats her score. So we need to find a way to calculate that.

Let’s consider the different ways exactly one remaining non-winner could finish with a higher total than Kristin:

only John does

only Jennifer does

only Andrew does

only Elliot does

only Scott does

only Michael does

That’s six different possibilities. What is the probability of each? I’ll make a table to make this more visual.

beats Kristin

John

Jennifer

Andrew

Elliot

Scott

Michael

Probability

John only

22%

78%

78%

78%

78%

78%

6.35%

Jennifer only

78%

22%

78%

78%

78%

78%

6.35%

Andrew only

78%

78%

22%

78%

78%

78%

6.35%

Elliot only

78%

78%

78%

22%

78%

78%

6.35%

Scott only

78%

78%

78%

78%

22%

78%

6.35%

Michael only

78%

78%

78%

78%

78%

22%

6.35%

Total

38.1%

Each of these combinations has a 6.35% chance of occurring.

When we want to calculate the probability that ANY combination happens, we add the probabilities together. Thus, using our assumptions, there is a 38.1% chance – six times 6.35% – that one and only one remaining player will pass Kristin’s score.

Combine this with the 22.5% probability that no one beats her and we get odds of around 60% for Kristin to advance. (In my actual analysis, I put her at 62%, because I didn’t round the initial 78% percentage.)

The #2 spot: many more possibilities

Let’s keep rolling. In second place after two quarterfinals was Dan Feitel, whose score of 14,000 was better than 82% of all non-winning scores, historically.

The probability that no one beats him:

82% x 82% x 82% x 82% x 82% x 82% = 0.826 = 30.4%

The probability that exactly one player beats him, using the pattern we identified with Kristin:

82% x 82% x 82% x 82% x 82% x 18% x 6 = 0.825 x 0.181 x 6 = 40.0%

Already, Dan has around a 70% chance of advancing. But he’ll also advance if exactly two players pass him. How do we calculate the odds that this takes place?

Let’s set up another table to find the various possibilities:

Combo #

John

Jennifer

Andrew

Elliot

Scott

Michael

Probability

1

18%

18%

82%

82%

82%

82%

1.46%

2

18%

82%

18%

82%

82%

82%

1.46%

3

18%

82%

82%

18%

82%

82%

1.46%

4

18%

82%

82%

82%

18%

82%

1.46%

5

18%

82%

82%

82%

82%

18%

1.46%

6

82%

18%

18%

82%

82%

82%

1.46%

7

82%

18%

82%

18%

82%

82%

1.46%

8

82%

18%

82%

82%

18%

82%

1.46%

9

82%

18%

82%

82%

82%

18%

1.46%

10

82%

82%

18%

18%

82%

82%

1.46%

11

82%

82%

18%

82%

18%

82%

1.46%

12

82%

82%

18%

82%

82%

18%

1.46%

13

82%

82%

82%

18%

18%

82%

1.46%

14

82%

82%

82%

18%

82%

18%

1.46%

15

82%

82%

82%

82%

18%

18%

1.46%

Total

22.0%

I count 15 different combinations of two players. That means the probability that exactly two players beat Dan is

82% x 82% x 82% x 82% x 18% x 18% x 15 = 0.824 x 0.182 x 15 = 22.0%

Let’s line these up and see if there’s a pattern.

Outcome

Worse

Better

Combinations

Total

Zero players better

0.826

x

0.180

x

1

30.4%

One player better

0.825

x

0.181

x

6

40.0%

Two players better

0.824

x

0.182

x

15

22.0%

Odds of advancing

92.4%

This looks promising: the exponents for the “worse” and “better” probabilities always add up to six, because all future players must finish either worse or better than our chosen player.

Have you figured out how we populate the “combinations” column? That’s right: in every case, we’re finding a “combination of n choose k” possibilities, where n is the total number of players remaining (six), and k is the number of players who finish higher than Dan in that particular scenario.

Recall that the number of combinations when you choose k objects from a population of n is

nCk

=

n!

k! x (n–k)!

Therefore:

6C0

=

6!

=

6!

=

1

0! x (6-0)!

1 x 6!

6C1

=

6!

=

6!

=

6 x 5!

=

6

1! x (6-1)!

1 x 5!

5!

6C2

=

6!

=

6!

=

6 x 5 x 4!

=

30

=

15

2! x (6-2)!

2 x 4!

2 x 4!

2

One of the neat things about combinations is it actually doesn’t matter whether I’m looking at the number of players who finish better than Dan, or the number of players who finish worse than Dan. If I were to instead find the number of combinations in which exactly five players finish with a lower total – 6 choose 5, in other words – I’ll get:

6C5

=

6!

=

6!

=

6 x 5!

=

6

5! x (6-5)!

5! x 1

5!

…which is exactly the same as 6C1 (six choose one), the number of combinations in which exactly one player finishes higher.

The #1 seed: an exercise

Time to put to work what you’ve learned so far.

After QF #2, Vaughn Winchell led the wild card pack with 16,599, a total higher than 95% of past non-winning scores.

Compared with Dan, Vaughn has one additional fallback to still qualify for the next round: if exactly three of the remaining non-winning players finish with a higher score.

Even without this option, Vaughn already has a 99.8% chance of moving on. But what should I put in the green cells to make this complete?

I’d never say that a player who hasn’t yet clinched a wild-card spot has a 100% chance of doing so – there’s always a possibility he’ll get bumped – but this is pretty much a lock.

If you’re interested in another, try this: in the 2004 Tournament of Champions, I sat in third place on the wild-card standings after 3 quarterfinal matches (meaning there were four players remaining who could beat me).

Assuming a player has a 40% chance of finishing below my score of 3,500, what were the odds that I’d advance?

I’ll fill out the other slots just to show that all of the possibilities add up to 100%. Note that we need to calculate a different number of combinations for each possibility, because there are only 4 players left!

Outcome

Worse

Better

Combinations

Total

Zero players better

0.404

x

0.600

x

1

2.6%

One player better

0.403

x

0.601

x

4

15.4%

Two players better

0.402

x

0.602

x

6

34.6%

Three players better

0.401

x

0.603

x

4

34.6%

Four players better

0.400

x

0.604

x

1

13.0%

Odds of advancing

18.0%

In reality, exactly two players finished higher than I did – unsurprising, given the odds. (The total probabilities add up to 100.2% due to rounding.)

Potential issues

Second, as I’ve pointed out in the past, Final Jeopardy! plays an outsized factor in determining wild cards, and there is probably an underlying correlation missing from the math above. The Final clue in this year’s QF #4 played much, much easier than the others, for example; Elliot Yates, however, didn’t have enough money to make anything of it.

Because the assignment of FJ! clues is random, I feel safe in ignoring any effects from the vagaries of their difficulty.

What’s next?

In my concluding post in this series – which I’ll certainly publish before the College Championship tapes in the new year – I’ll share the model behind my Wild Card Wagering Calculator (WC2), which takes into account your confidence level in the category.

For example, if you’ve got 8,000 heading into Final Jeopardy! in a ToC quarterfinal, and you’ll probably finish second in the game, you should wager as follows: